/*
 * Copyright (c) 2009, 2013, Oracle and/or its affiliates. All rights reserved.
 * Copyright 2009 Google Inc.  All Rights Reserved.
 * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
 *
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 *
 *
 */

package java.util;

/**
 * This is a near duplicate of {@link TimSort}, modified for use with
 * arrays of objects that implement {@link Comparable}, instead of using
 * explicit comparators.
 *
 * <p>If you are using an optimizing VM, you may find that ComparableTimSort
 * offers no performance benefit over TimSort in conjunction with a
 * comparator that simply returns {@code ((Comparable)first).compareTo(Second)}.
 * If this is the case, you are better off deleting ComparableTimSort to
 * eliminate the code duplication.  (See Arrays.java for details.)
 *
 * @author Josh Bloch
 */
class ComparableTimSort {

  /**
   * This is the minimum sized sequence that will be merged.  Shorter
   * sequences will be lengthened by calling binarySort.  If the entire
   * array is less than this length, no merges will be performed.
   *
   * This constant should be a power of two.  It was 64 in Tim Peter's C
   * implementation, but 32 was empirically determined to work better in
   * this implementation.  In the unlikely event that you set this constant
   * to be a number that's not a power of two, you'll need to change the
   * {@link #minRunLength} computation.
   *
   * If you decrease this constant, you must change the stackLen
   * computation in the TimSort constructor, or you risk an
   * ArrayOutOfBounds exception.  See listsort.txt for a discussion
   * of the minimum stack length required as a function of the length
   * of the array being sorted and the minimum merge sequence length.
   */
  private static final int MIN_MERGE = 32;

  /**
   * The array being sorted.
   */
  private final Object[] a;

  /**
   * When we get into galloping mode, we stay there until both runs win less
   * often than MIN_GALLOP consecutive times.
   */
  private static final int MIN_GALLOP = 7;

  /**
   * This controls when we get *into* galloping mode.  It is initialized
   * to MIN_GALLOP.  The mergeLo and mergeHi methods nudge it higher for
   * random data, and lower for highly structured data.
   */
  private int minGallop = MIN_GALLOP;

  /**
   * Maximum initial size of tmp array, which is used for merging.  The array
   * can grow to accommodate demand.
   *
   * Unlike Tim's original C version, we do not allocate this much storage
   * when sorting smaller arrays.  This change was required for performance.
   */
  private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

  /**
   * Temp storage for merges. A workspace array may optionally be
   * provided in constructor, and if so will be used as long as it
   * is big enough.
   */
  private Object[] tmp;
  private int tmpBase; // base of tmp array slice
  private int tmpLen;  // length of tmp array slice

  /**
   * A stack of pending runs yet to be merged.  Run i starts at
   * address base[i] and extends for len[i] elements.  It's always
   * true (so long as the indices are in bounds) that:
   *
   * runBase[i] + runLen[i] == runBase[i + 1]
   *
   * so we could cut the storage for this, but it's a minor amount,
   * and keeping all the info explicit simplifies the code.
   */
  private int stackSize = 0;  // Number of pending runs on stack
  private final int[] runBase;
  private final int[] runLen;

  /**
   * Creates a TimSort instance to maintain the state of an ongoing sort.
   *
   * @param a the array to be sorted
   * @param work a workspace array (slice)
   * @param workBase origin of usable space in work array
   * @param workLen usable size of work array
   */
  private ComparableTimSort(Object[] a, Object[] work, int workBase, int workLen) {
    this.a = a;

    // Allocate temp storage (which may be increased later if necessary)
    int len = a.length;
    int tlen = (len < 2 * INITIAL_TMP_STORAGE_LENGTH) ?
        len >>> 1 : INITIAL_TMP_STORAGE_LENGTH;
    if (work == null || workLen < tlen || workBase + tlen > work.length) {
      tmp = new Object[tlen];
      tmpBase = 0;
      tmpLen = tlen;
    } else {
      tmp = work;
      tmpBase = workBase;
      tmpLen = workLen;
    }

        /*
         * Allocate runs-to-be-merged stack (which cannot be expanded).  The
         * stack length requirements are described in listsort.txt.  The C
         * version always uses the same stack length (85), but this was
         * measured to be too expensive when sorting "mid-sized" arrays (e.g.,
         * 100 elements) in Java.  Therefore, we use smaller (but sufficiently
         * large) stack lengths for smaller arrays.  The "magic numbers" in the
         * computation below must be changed if MIN_MERGE is decreased.  See
         * the MIN_MERGE declaration above for more information.
         * The maximum value of 49 allows for an array up to length
         * Integer.MAX_VALUE-4, if array is filled by the worst case stack size
         * increasing scenario. More explanations are given in section 4 of:
         * http://envisage-project.eu/wp-content/uploads/2015/02/sorting.pdf
         */
    int stackLen = (len < 120 ? 5 :
        len < 1542 ? 10 :
            len < 119151 ? 24 : 49);
    runBase = new int[stackLen];
    runLen = new int[stackLen];
  }

    /*
     * The next method (package private and static) constitutes the
     * entire API of this class.
     */

  /**
   * Sorts the given range, using the given workspace array slice
   * for temp storage when possible. This method is designed to be
   * invoked from public methods (in class Arrays) after performing
   * any necessary array bounds checks and expanding parameters into
   * the required forms.
   *
   * @param a the array to be sorted
   * @param lo the index of the first element, inclusive, to be sorted
   * @param hi the index of the last element, exclusive, to be sorted
   * @param work a workspace array (slice)
   * @param workBase origin of usable space in work array
   * @param workLen usable size of work array
   * @since 1.8
   */
  static void sort(Object[] a, int lo, int hi, Object[] work, int workBase, int workLen) {
    assert a != null && lo >= 0 && lo <= hi && hi <= a.length;

    int nRemaining = hi - lo;
    if (nRemaining < 2) {
      return;  // Arrays of size 0 and 1 are always sorted
    }

    // If array is small, do a "mini-TimSort" with no merges
    if (nRemaining < MIN_MERGE) {
      int initRunLen = countRunAndMakeAscending(a, lo, hi);
      binarySort(a, lo, hi, lo + initRunLen);
      return;
    }

    /**
     * March over the array once, left to right, finding natural runs,
     * extending short natural runs to minRun elements, and merging runs
     * to maintain stack invariant.
     */
    ComparableTimSort ts = new ComparableTimSort(a, work, workBase, workLen);
    int minRun = minRunLength(nRemaining);
    do {
      // Identify next run
      int runLen = countRunAndMakeAscending(a, lo, hi);

      // If run is short, extend to min(minRun, nRemaining)
      if (runLen < minRun) {
        int force = nRemaining <= minRun ? nRemaining : minRun;
        binarySort(a, lo, lo + force, lo + runLen);
        runLen = force;
      }

      // Push run onto pending-run stack, and maybe merge
      ts.pushRun(lo, runLen);
      ts.mergeCollapse();

      // Advance to find next run
      lo += runLen;
      nRemaining -= runLen;
    } while (nRemaining != 0);

    // Merge all remaining runs to complete sort
    assert lo == hi;
    ts.mergeForceCollapse();
    assert ts.stackSize == 1;
  }

  /**
   * Sorts the specified portion of the specified array using a binary
   * insertion sort.  This is the best method for sorting small numbers
   * of elements.  It requires O(n log n) compares, but O(n^2) data
   * movement (worst case).
   *
   * If the initial part of the specified range is already sorted,
   * this method can take advantage of it: the method assumes that the
   * elements from index {@code lo}, inclusive, to {@code start},
   * exclusive are already sorted.
   *
   * @param a the array in which a range is to be sorted
   * @param lo the index of the first element in the range to be sorted
   * @param hi the index after the last element in the range to be sorted
   * @param start the index of the first element in the range that is not already known to be sorted
   * ({@code lo <= start <= hi})
   */
  @SuppressWarnings({"fallthrough", "rawtypes", "unchecked"})
  private static void binarySort(Object[] a, int lo, int hi, int start) {
    assert lo <= start && start <= hi;
    if (start == lo) {
      start++;
    }
    for (; start < hi; start++) {
      Comparable pivot = (Comparable) a[start];

      // Set left (and right) to the index where a[start] (pivot) belongs
      int left = lo;
      int right = start;
      assert left <= right;
            /*
             * Invariants:
             *   pivot >= all in [lo, left).
             *   pivot <  all in [right, start).
             */
      while (left < right) {
        int mid = (left + right) >>> 1;
        if (pivot.compareTo(a[mid]) < 0) {
          right = mid;
        } else {
          left = mid + 1;
        }
      }
      assert left == right;

            /*
             * The invariants still hold: pivot >= all in [lo, left) and
             * pivot < all in [left, start), so pivot belongs at left.  Note
             * that if there are elements equal to pivot, left points to the
             * first slot after them -- that's why this sort is stable.
             * Slide elements over to make room for pivot.
             */
      int n = start - left;  // The number of elements to move
      // Switch is just an optimization for arraycopy in default case
      switch (n) {
        case 2:
          a[left + 2] = a[left + 1];
        case 1:
          a[left + 1] = a[left];
          break;
        default:
          System.arraycopy(a, left, a, left + 1, n);
      }
      a[left] = pivot;
    }
  }

  /**
   * Returns the length of the run beginning at the specified position in
   * the specified array and reverses the run if it is descending (ensuring
   * that the run will always be ascending when the method returns).
   *
   * A run is the longest ascending sequence with:
   *
   * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
   *
   * or the longest descending sequence with:
   *
   * a[lo] >  a[lo + 1] >  a[lo + 2] >  ...
   *
   * For its intended use in a stable mergesort, the strictness of the
   * definition of "descending" is needed so that the call can safely
   * reverse a descending sequence without violating stability.
   *
   * @param a the array in which a run is to be counted and possibly reversed
   * @param lo index of the first element in the run
   * @param hi index after the last element that may be contained in the run. It is required that
   * {@code lo < hi}.
   * @return the length of the run beginning at the specified position in the specified array
   */
  @SuppressWarnings({"unchecked", "rawtypes"})
  private static int countRunAndMakeAscending(Object[] a, int lo, int hi) {
    assert lo < hi;
    int runHi = lo + 1;
    if (runHi == hi) {
      return 1;
    }

    // Find end of run, and reverse range if descending
    if (((Comparable) a[runHi++]).compareTo(a[lo]) < 0) { // Descending
      while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) < 0) {
        runHi++;
      }
      reverseRange(a, lo, runHi);
    } else {                              // Ascending
      while (runHi < hi && ((Comparable) a[runHi]).compareTo(a[runHi - 1]) >= 0) {
        runHi++;
      }
    }

    return runHi - lo;
  }

  /**
   * Reverse the specified range of the specified array.
   *
   * @param a the array in which a range is to be reversed
   * @param lo the index of the first element in the range to be reversed
   * @param hi the index after the last element in the range to be reversed
   */
  private static void reverseRange(Object[] a, int lo, int hi) {
    hi--;
    while (lo < hi) {
      Object t = a[lo];
      a[lo++] = a[hi];
      a[hi--] = t;
    }
  }

  /**
   * Returns the minimum acceptable run length for an array of the specified
   * length. Natural runs shorter than this will be extended with
   * {@link #binarySort}.
   *
   * Roughly speaking, the computation is:
   *
   * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
   * Else if n is an exact power of 2, return MIN_MERGE/2.
   * Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such that n/k
   * is close to, but strictly less than, an exact power of 2.
   *
   * For the rationale, see listsort.txt.
   *
   * @param n the length of the array to be sorted
   * @return the length of the minimum run to be merged
   */
  private static int minRunLength(int n) {
    assert n >= 0;
    int r = 0;      // Becomes 1 if any 1 bits are shifted off
    while (n >= MIN_MERGE) {
      r |= (n & 1);
      n >>= 1;
    }
    return n + r;
  }

  /**
   * Pushes the specified run onto the pending-run stack.
   *
   * @param runBase index of the first element in the run
   * @param runLen the number of elements in the run
   */
  private void pushRun(int runBase, int runLen) {
    this.runBase[stackSize] = runBase;
    this.runLen[stackSize] = runLen;
    stackSize++;
  }

  /**
   * Examines the stack of runs waiting to be merged and merges adjacent runs
   * until the stack invariants are reestablished:
   *
   * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1]
   * 2. runLen[i - 2] > runLen[i - 1]
   *
   * This method is called each time a new run is pushed onto the stack,
   * so the invariants are guaranteed to hold for i < stackSize upon
   * entry to the method.
   */
  private void mergeCollapse() {
    while (stackSize > 1) {
      int n = stackSize - 2;
      if (n > 0 && runLen[n - 1] <= runLen[n] + runLen[n + 1]) {
        if (runLen[n - 1] < runLen[n + 1]) {
          n--;
        }
        mergeAt(n);
      } else if (runLen[n] <= runLen[n + 1]) {
        mergeAt(n);
      } else {
        break; // Invariant is established
      }
    }
  }

  /**
   * Merges all runs on the stack until only one remains.  This method is
   * called once, to complete the sort.
   */
  private void mergeForceCollapse() {
    while (stackSize > 1) {
      int n = stackSize - 2;
      if (n > 0 && runLen[n - 1] < runLen[n + 1]) {
        n--;
      }
      mergeAt(n);
    }
  }

  /**
   * Merges the two runs at stack indices i and i+1.  Run i must be
   * the penultimate or antepenultimate run on the stack.  In other words,
   * i must be equal to stackSize-2 or stackSize-3.
   *
   * @param i stack index of the first of the two runs to merge
   */
  @SuppressWarnings("unchecked")
  private void mergeAt(int i) {
    assert stackSize >= 2;
    assert i >= 0;
    assert i == stackSize - 2 || i == stackSize - 3;

    int base1 = runBase[i];
    int len1 = runLen[i];
    int base2 = runBase[i + 1];
    int len2 = runLen[i + 1];
    assert len1 > 0 && len2 > 0;
    assert base1 + len1 == base2;

        /*
         * Record the length of the combined runs; if i is the 3rd-last
         * run now, also slide over the last run (which isn't involved
         * in this merge).  The current run (i+1) goes away in any case.
         */
    runLen[i] = len1 + len2;
    if (i == stackSize - 3) {
      runBase[i + 1] = runBase[i + 2];
      runLen[i + 1] = runLen[i + 2];
    }
    stackSize--;

        /*
         * Find where the first element of run2 goes in run1. Prior elements
         * in run1 can be ignored (because they're already in place).
         */
    int k = gallopRight((Comparable<Object>) a[base2], a, base1, len1, 0);
    assert k >= 0;
    base1 += k;
    len1 -= k;
    if (len1 == 0) {
      return;
    }

        /*
         * Find where the last element of run1 goes in run2. Subsequent elements
         * in run2 can be ignored (because they're already in place).
         */
    len2 = gallopLeft((Comparable<Object>) a[base1 + len1 - 1], a,
        base2, len2, len2 - 1);
    assert len2 >= 0;
    if (len2 == 0) {
      return;
    }

    // Merge remaining runs, using tmp array with min(len1, len2) elements
    if (len1 <= len2) {
      mergeLo(base1, len1, base2, len2);
    } else {
      mergeHi(base1, len1, base2, len2);
    }
  }

  /**
   * Locates the position at which to insert the specified key into the
   * specified sorted range; if the range contains an element equal to key,
   * returns the index of the leftmost equal element.
   *
   * @param key the key whose insertion point to search for
   * @param a the array in which to search
   * @param base the index of the first element in the range
   * @param len the length of the range; must be > 0
   * @param hint the index at which to begin the search, 0 <= hint < n. The closer hint is to the
   * result, the faster this method will run.
   * @return the int k,  0 <= k <= n such that a[b + k - 1] < key <= a[b + k], pretending that a[b -
   * 1] is minus infinity and a[b + n] is infinity. In other words, key belongs at index b + k; or
   * in other words, the first k elements of a should precede key, and the last n - k should follow
   * it.
   */
  private static int gallopLeft(Comparable<Object> key, Object[] a,
      int base, int len, int hint) {
    assert len > 0 && hint >= 0 && hint < len;

    int lastOfs = 0;
    int ofs = 1;
    if (key.compareTo(a[base + hint]) > 0) {
      // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
      int maxOfs = len - hint;
      while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) > 0) {
        lastOfs = ofs;
        ofs = (ofs << 1) + 1;
        if (ofs <= 0)   // int overflow
        {
          ofs = maxOfs;
        }
      }
      if (ofs > maxOfs) {
        ofs = maxOfs;
      }

      // Make offsets relative to base
      lastOfs += hint;
      ofs += hint;
    } else { // key <= a[base + hint]
      // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
      final int maxOfs = hint + 1;
      while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) <= 0) {
        lastOfs = ofs;
        ofs = (ofs << 1) + 1;
        if (ofs <= 0)   // int overflow
        {
          ofs = maxOfs;
        }
      }
      if (ofs > maxOfs) {
        ofs = maxOfs;
      }

      // Make offsets relative to base
      int tmp = lastOfs;
      lastOfs = hint - ofs;
      ofs = hint - tmp;
    }
    assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere
         * to the right of lastOfs but no farther right than ofs.  Do a binary
         * search, with invariant a[base + lastOfs - 1] < key <= a[base + ofs].
         */
    lastOfs++;
    while (lastOfs < ofs) {
      int m = lastOfs + ((ofs - lastOfs) >>> 1);

      if (key.compareTo(a[base + m]) > 0) {
        lastOfs = m + 1;  // a[base + m] < key
      } else {
        ofs = m;          // key <= a[base + m]
      }
    }
    assert lastOfs == ofs;    // so a[base + ofs - 1] < key <= a[base + ofs]
    return ofs;
  }

  /**
   * Like gallopLeft, except that if the range contains an element equal to
   * key, gallopRight returns the index after the rightmost equal element.
   *
   * @param key the key whose insertion point to search for
   * @param a the array in which to search
   * @param base the index of the first element in the range
   * @param len the length of the range; must be > 0
   * @param hint the index at which to begin the search, 0 <= hint < n. The closer hint is to the
   * result, the faster this method will run.
   * @return the int k,  0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
   */
  private static int gallopRight(Comparable<Object> key, Object[] a,
      int base, int len, int hint) {
    assert len > 0 && hint >= 0 && hint < len;

    int ofs = 1;
    int lastOfs = 0;
    if (key.compareTo(a[base + hint]) < 0) {
      // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
      int maxOfs = hint + 1;
      while (ofs < maxOfs && key.compareTo(a[base + hint - ofs]) < 0) {
        lastOfs = ofs;
        ofs = (ofs << 1) + 1;
        if (ofs <= 0)   // int overflow
        {
          ofs = maxOfs;
        }
      }
      if (ofs > maxOfs) {
        ofs = maxOfs;
      }

      // Make offsets relative to b
      int tmp = lastOfs;
      lastOfs = hint - ofs;
      ofs = hint - tmp;
    } else { // a[b + hint] <= key
      // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
      int maxOfs = len - hint;
      while (ofs < maxOfs && key.compareTo(a[base + hint + ofs]) >= 0) {
        lastOfs = ofs;
        ofs = (ofs << 1) + 1;
        if (ofs <= 0)   // int overflow
        {
          ofs = maxOfs;
        }
      }
      if (ofs > maxOfs) {
        ofs = maxOfs;
      }

      // Make offsets relative to b
      lastOfs += hint;
      ofs += hint;
    }
    assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
         * the right of lastOfs but no farther right than ofs.  Do a binary
         * search, with invariant a[b + lastOfs - 1] <= key < a[b + ofs].
         */
    lastOfs++;
    while (lastOfs < ofs) {
      int m = lastOfs + ((ofs - lastOfs) >>> 1);

      if (key.compareTo(a[base + m]) < 0) {
        ofs = m;          // key < a[b + m]
      } else {
        lastOfs = m + 1;  // a[b + m] <= key
      }
    }
    assert lastOfs == ofs;    // so a[b + ofs - 1] <= key < a[b + ofs]
    return ofs;
  }

  /**
   * Merges two adjacent runs in place, in a stable fashion.  The first
   * element of the first run must be greater than the first element of the
   * second run (a[base1] > a[base2]), and the last element of the first run
   * (a[base1 + len1-1]) must be greater than all elements of the second run.
   *
   * For performance, this method should be called only when len1 <= len2;
   * its twin, mergeHi should be called if len1 >= len2.  (Either method
   * may be called if len1 == len2.)
   *
   * @param base1 index of first element in first run to be merged
   * @param len1 length of first run to be merged (must be > 0)
   * @param base2 index of first element in second run to be merged (must be aBase + aLen)
   * @param len2 length of second run to be merged (must be > 0)
   */
  @SuppressWarnings({"unchecked", "rawtypes"})
  private void mergeLo(int base1, int len1, int base2, int len2) {
    assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

    // Copy first run into temp array
    Object[] a = this.a; // For performance
    Object[] tmp = ensureCapacity(len1);

    int cursor1 = tmpBase; // Indexes into tmp array
    int cursor2 = base2;   // Indexes int a
    int dest = base1;      // Indexes int a
    System.arraycopy(a, base1, tmp, cursor1, len1);

    // Move first element of second run and deal with degenerate cases
    a[dest++] = a[cursor2++];
    if (--len2 == 0) {
      System.arraycopy(tmp, cursor1, a, dest, len1);
      return;
    }
    if (len1 == 1) {
      System.arraycopy(a, cursor2, a, dest, len2);
      a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
      return;
    }

    int minGallop = this.minGallop;  // Use local variable for performance
    outer:
    while (true) {
      int count1 = 0; // Number of times in a row that first run won
      int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run starts
             * winning consistently.
             */
      do {
        assert len1 > 1 && len2 > 0;
        if (((Comparable) a[cursor2]).compareTo(tmp[cursor1]) < 0) {
          a[dest++] = a[cursor2++];
          count2++;
          count1 = 0;
          if (--len2 == 0) {
            break outer;
          }
        } else {
          a[dest++] = tmp[cursor1++];
          count1++;
          count2 = 0;
          if (--len1 == 1) {
            break outer;
          }
        }
      } while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
      do {
        assert len1 > 1 && len2 > 0;
        count1 = gallopRight((Comparable) a[cursor2], tmp, cursor1, len1, 0);
        if (count1 != 0) {
          System.arraycopy(tmp, cursor1, a, dest, count1);
          dest += count1;
          cursor1 += count1;
          len1 -= count1;
          if (len1 <= 1)  // len1 == 1 || len1 == 0
          {
            break outer;
          }
        }
        a[dest++] = a[cursor2++];
        if (--len2 == 0) {
          break outer;
        }

        count2 = gallopLeft((Comparable) tmp[cursor1], a, cursor2, len2, 0);
        if (count2 != 0) {
          System.arraycopy(a, cursor2, a, dest, count2);
          dest += count2;
          cursor2 += count2;
          len2 -= count2;
          if (len2 == 0) {
            break outer;
          }
        }
        a[dest++] = tmp[cursor1++];
        if (--len1 == 1) {
          break outer;
        }
        minGallop--;
      } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
      if (minGallop < 0) {
        minGallop = 0;
      }
      minGallop += 2;  // Penalize for leaving gallop mode
    }  // End of "outer" loop
    this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

    if (len1 == 1) {
      assert len2 > 0;
      System.arraycopy(a, cursor2, a, dest, len2);
      a[dest + len2] = tmp[cursor1]; //  Last elt of run 1 to end of merge
    } else if (len1 == 0) {
      throw new IllegalArgumentException(
          "Comparison method violates its general contract!");
    } else {
      assert len2 == 0;
      assert len1 > 1;
      System.arraycopy(tmp, cursor1, a, dest, len1);
    }
  }

  /**
   * Like mergeLo, except that this method should be called only if
   * len1 >= len2; mergeLo should be called if len1 <= len2.  (Either method
   * may be called if len1 == len2.)
   *
   * @param base1 index of first element in first run to be merged
   * @param len1 length of first run to be merged (must be > 0)
   * @param base2 index of first element in second run to be merged (must be aBase + aLen)
   * @param len2 length of second run to be merged (must be > 0)
   */
  @SuppressWarnings({"unchecked", "rawtypes"})
  private void mergeHi(int base1, int len1, int base2, int len2) {
    assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

    // Copy second run into temp array
    Object[] a = this.a; // For performance
    Object[] tmp = ensureCapacity(len2);
    int tmpBase = this.tmpBase;
    System.arraycopy(a, base2, tmp, tmpBase, len2);

    int cursor1 = base1 + len1 - 1;  // Indexes into a
    int cursor2 = tmpBase + len2 - 1; // Indexes into tmp array
    int dest = base2 + len2 - 1;     // Indexes into a

    // Move last element of first run and deal with degenerate cases
    a[dest--] = a[cursor1--];
    if (--len1 == 0) {
      System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
      return;
    }
    if (len2 == 1) {
      dest -= len1;
      cursor1 -= len1;
      System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
      a[dest] = tmp[cursor2];
      return;
    }

    int minGallop = this.minGallop;  // Use local variable for performance
    outer:
    while (true) {
      int count1 = 0; // Number of times in a row that first run won
      int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run
             * appears to win consistently.
             */
      do {
        assert len1 > 0 && len2 > 1;
        if (((Comparable) tmp[cursor2]).compareTo(a[cursor1]) < 0) {
          a[dest--] = a[cursor1--];
          count1++;
          count2 = 0;
          if (--len1 == 0) {
            break outer;
          }
        } else {
          a[dest--] = tmp[cursor2--];
          count2++;
          count1 = 0;
          if (--len2 == 1) {
            break outer;
          }
        }
      } while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a
             * huge win. So try that, and continue galloping until (if ever)
             * neither run appears to be winning consistently anymore.
             */
      do {
        assert len1 > 0 && len2 > 1;
        count1 = len1 - gallopRight((Comparable) tmp[cursor2], a, base1, len1, len1 - 1);
        if (count1 != 0) {
          dest -= count1;
          cursor1 -= count1;
          len1 -= count1;
          System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
          if (len1 == 0) {
            break outer;
          }
        }
        a[dest--] = tmp[cursor2--];
        if (--len2 == 1) {
          break outer;
        }

        count2 = len2 - gallopLeft((Comparable) a[cursor1], tmp, tmpBase, len2, len2 - 1);
        if (count2 != 0) {
          dest -= count2;
          cursor2 -= count2;
          len2 -= count2;
          System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
          if (len2 <= 1) {
            break outer; // len2 == 1 || len2 == 0
          }
        }
        a[dest--] = a[cursor1--];
        if (--len1 == 0) {
          break outer;
        }
        minGallop--;
      } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
      if (minGallop < 0) {
        minGallop = 0;
      }
      minGallop += 2;  // Penalize for leaving gallop mode
    }  // End of "outer" loop
    this.minGallop = minGallop < 1 ? 1 : minGallop;  // Write back to field

    if (len2 == 1) {
      assert len1 > 0;
      dest -= len1;
      cursor1 -= len1;
      System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
      a[dest] = tmp[cursor2];  // Move first elt of run2 to front of merge
    } else if (len2 == 0) {
      throw new IllegalArgumentException(
          "Comparison method violates its general contract!");
    } else {
      assert len1 == 0;
      assert len2 > 0;
      System.arraycopy(tmp, tmpBase, a, dest - (len2 - 1), len2);
    }
  }

  /**
   * Ensures that the external array tmp has at least the specified
   * number of elements, increasing its size if necessary.  The size
   * increases exponentially to ensure amortized linear time complexity.
   *
   * @param minCapacity the minimum required capacity of the tmp array
   * @return tmp, whether or not it grew
   */
  private Object[] ensureCapacity(int minCapacity) {
    if (tmpLen < minCapacity) {
      // Compute smallest power of 2 > minCapacity
      int newSize = minCapacity;
      newSize |= newSize >> 1;
      newSize |= newSize >> 2;
      newSize |= newSize >> 4;
      newSize |= newSize >> 8;
      newSize |= newSize >> 16;
      newSize++;

      if (newSize < 0) // Not bloody likely!
      {
        newSize = minCapacity;
      } else {
        newSize = Math.min(newSize, a.length >>> 1);
      }

      @SuppressWarnings({"unchecked", "UnnecessaryLocalVariable"})
      Object[] newArray = new Object[newSize];
      tmp = newArray;
      tmpLen = newSize;
      tmpBase = 0;
    }
    return tmp;
  }

}
